2–74. Integration techniques Use the methods introduced in Sec...

Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

65. ∫ (from 0 to 1) dy/((y + 1)(y² + 1))

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to evaluate the integral, and we're given the integral from 0 to 1 of DX divided by the quantity of the quantity of X plus 2 in quantity multiplied by the quantity of X2 + 4 in quantity. So the first step is to rewrite our integram into multiple fractions using partial fractions. So we call for a partial fraction, that 1 divided by the quantity of the quantity of X plus 2. In quantity multiplied by the quantity of X2 + 4 in quantity is going to be equal to a divided by the quantity of X plus 2 in quantity plus the quantity of B X plus C in quantity divided by the quantity of X2 + 4 in quantity. The next step is to multiply both sides of the equation by the denominator on the left hand side to remove all the fractions. In doing so, we will get 1 is equal to a multiplied by the quantity of X2 + 4 in quantity. Plus the quantity of B X plus C in quantity multiplied by the quantity of X plus 2 in quantity. And now multiplying everything out, we'll have that 1 is equal to AX2 + 4 A. Plus, B X squared. Plus 2 BX. Plus DX. Plus 2 C. and collecting like terms, we'll have that 1 is equal to the quantity of A plus B in quantity multiplied by X2. Plus, the quantity of 2B. Plus D in quantity multiplied by X. Plus 4 A + 2 C. And so now we just need to match up the coefficients of like powers of X on both sides of the equation. So, for the quadratic term, we will have that A plus B is going to be equal to 0, since we have no quadratic term on the left hand side. For the linear term, we will have that 2B plus C is going to be equal to 0, and for the constant term, we will have that 4A plus 2 C is equal to 1. Now, from the first equation, we can subtract A from both sides of the equation, and that means that B is going to be equal to minus A. So we can substitute this into the second equation, and we will have that minus 2 A. Plus C is equal to 0. And if we solve this for C, we'll see that C is equal to 2 A. And so substituting this value for C into the last equation, we will have that. Or a Plus 2 multiplied by 2 A. is equal to one. And to solve this for A, we'll just add the terms together on the left hand side, and we will get that 8A is equal to 1. And divide both sides by 8, we'll get that A is equal to 1 divided by 8. And that means that D is going to be equal to 2 divided by 8, which is equal to 1 divided by 4. And that means that B. is going to be equal to -1 divided by 8. So now we found all three coefficients, we can rewrite our integral, and that is the interval. From 0 to 1 of DX divided by the quantity of the quantity of X plus 2, in quantity multiplied by the quantity of X2 + 4 in quantity. And this is going to be equal to. The integral From 0 to 1 of the quantity of 1 divided by 8 in quantity, divided by the quantity of X plus 2 in quantity, DX. Plus, the integral from 0 to 1. Of the quantity of minus 1 divided by 8, multiplied by X. Plus 1 divided by or in quantity, divided by the quantity of X2 plus 4 in quantity DX. So let's evaluate each one of these integrals individually. So, if we look at the interval, From 0 to 1. Of the quantity of 1 divided by 8 in quantity, divided by the quantity of X plus 2 in quantity DX. That this is going to be equal to, and we can here we can perform the integration, so we'll hold the constant factor of 1 divided by 8 out front. So we'll have 1 divided by 8, and this gets multiplied by, and here we'll be integrating 1 divided by the quantity of X + 2 in quantity with respect to X, and recall that's going to give us the natural log of the absolute value of the quantity of X plus 2. In quantity, and this needs to be evaluated from 0 to 1. So substituting in the limits of integration, this is going to be equal to 1 divided by 8, multiplied by the quantity of the natural log of the absolute value of the quantity of 1 + 2 in quantity minus the natural log of the absolute value of the quantity of 0 + 2 in quantity. And simplifying this expression, this is going to be equal to 1 divided by 8, multiplied by the natural log of the quantity of 3 divided by 2 in quantity. And so here we've dropped the absolute value signs since the argument inside the absolute value function is going to be positive for both of those um absolute value functions, and here we just combine the two natural log functions using the properties of logarithms. So, next, we're going to look at the second interval, which is going to be the interval from 0 to 1 of quantity of minus 1 divided by 8, multiplied by X. Plus 1 divided by 4 in quantity, divided by the quantity of X2 + 4 in quantity DX. And what we want to do is to rewrite this as two separate integrals. So this is going to be equal to minus 1 divided by 8, multiplied by the integral from 0 to 1 of X, divided by the quantity of. X2 + 4 in quantity, D X. Plus 1 divided by 4, multiplied by the integral from 0 to 1 of the quantity of 1 divided by the quantity of X squared plus 4 in quantity DX. Now, for the first integral here, we're gonna have to make a substitution. So, we'll say U equal to X2 + 4. That means that DU is going to be equal to 2 XDX. So making this substitution, this is going to be equal to. Minus 1 divided by 16, multiplied by the integral, and here we have to change our limits of integration as we're changing our variables. So when X is equal to 0, U is going to be equal to 02 + 4, so that's going to be equal to 4. And for our upper limited integration, when X is equals 1, U is going to be equal to 12 + 4, which is 5. And then for the integram, we're going to have 1 divided by U. DU And then we'll have plus, and for a second interval, we do not need to do anything, so we'll just have 1 divided by 4, multiplied by the integral from 0 to 1 of 1, divided by the quantity of X squared, plus 4 in quantity DX. So now we can perform the integration. In doing so, this is going to be equal to. -1 divided by 16, multiplied by the quantity, and here we're going to be integrating 1 divided by you with respect to you, and that will give us the natural log of the absolute value of you. And in quantity, and that's gonna be evaluated from 4 to 5. Then we'll have plus. 1 divided by 4, and this is going to be multiplied by, and here we're integrating 1 divided by the quantity of X2 plus 4 in quantity with respect to X, and that's going to give us The quantity of 1 divided by 2 multiplied by the inverse tangent of the quantity of X divided by 2 in quantity, and this needs to be evaluated from 0 to 1. So substituting in the limits of integration, this is going to be equal to. -1 divided by 16, multiplied by the quantity of the natural log of 5, minus the natural log of 4 in quantity. Plus 1 divided by 8, multiplied by the quantity of the inverse tangent. Of the quantity of 1 divided by 2 in quantity minus the inverse tangent of the quantity of 0 divided by 2 in quantity. And so simplifying this expression, this is going to be equal to -1 divided by 16, multiplied by the natural log of the quantity of 5 divided by 4 in quantity. Plus 1 divided by 8, multiplied by the inverse tangent of the quantity of 1 divided by 2 in quantity, since the inverse tangent of 0 is going to be 0. And so now we just need to combine our results so that the integral from 0 to 1 of DX divided by the quantity of the quantity of X plus 2 in quantity, multiplied by the quantity of X squared plus 4 in quantity is going to be equal to. 1 divided by 8, multiplied by the natural log of the quantity of 3 divided by 2. In Quanti Minus 1 divided by 16. Multiplied by the natural log of the quantity of 5 divided by 4 in quantity. Plus 1 divided by 8, multiplied by the inverse tangent of the quantity of 1 divided by 2 in quantity. So now we just want to combine our natural functions, but we need to rewrite that first term, so that is going to be equal to. 1 divided by 16, multiplied by the natural log of the quantity of 9 divided by 4. In quantity, minus 1 divided by 16, multiplied by the natural log of the quantity of 5 divided by 4 in quantity, plus 1 divided by 8 multiplied by the inverse tangent. Of the quantity of 1 divided by 2 in quantity. And so combining our natural log functions, this is going to be equal to 1 divided by 16 multiplied by the natural log of quantity of 9 divided by 5 in quantity, plus 1 divided by 8 multiplied by the inverse tangent of the quantity of 1 divided by 2 in quantity. And this here is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.65. ∫ (from 0 to 1) dy/((y + 1)(y² + 1))

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2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
65. ∫ (from 0 to 1) dy/((y + 1)(y² + 1))